# Logarithm question for Algebra 2/Trig class

$$\frac{1}{2} \log(x+2)=2$$

I'm decently good at logarithms but this one seems to be tricky, when I did it myself I got a negative decimal as my answer but I'm not 100% confident in it, and I would really appreciate some help!

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Is your equation $(1/2)\log(x+2)=2$ or $1/(2\log(x+2))=2$? – David Mitra Jan 12 '13 at 19:10
Obtaining a negative value $x$ as a solution is fine, here, for the equation ${1\over 2\log(x+2)}=2$ (where $\log$ is the natural logarithm). Solving this, we obtain the solution $x=e^{1/4}-2$. This is negative; and no doubt you know you can't take the logarithm of a negative number. But we don't do that here: we take the logarithm of $x\color{maroon}{+2}$, which is positive for our $x$. And this solution does "work". It's always a good idea to check your solutions when solving equations. – David Mitra Jan 12 '13 at 19:16
The equation is (1/2)log(x+2)=2 David – user57055 Jan 12 '13 at 19:20
You shouldn't have obtained a negative solution then. (The solution is $e^4-2$ for the natural logarithm, and $10^4-2$ for the common logarithm.) – David Mitra Jan 12 '13 at 19:25
Yes it is a common logarithm, the problem on my homework is literally displayed without the parenthesis around the 1/2, that is why I posted my question as 1/2log(x+2)=2. One of questions I had a bit of trouble really piecing together. – user57055 Jan 12 '13 at 19:27

You have

$\frac{1}{2} \log(x+2)=2$

multiply both sides for 2

$\log(x+2)=4$

Now, I suppose the logarithm base is $e$ so, raise $e$ to both sides of the equation

$(x+2)=e^4$

so, $x=e^4-2$.

Similarly, if the base of the logarithm is 10, the answer is $x=10^4-2$

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Thank you so much, finally put all the pieces together and found out what I did wrong. The log base of 10 is already built into the word log on my TI-84+ so without the base of 10, the problem would be wrong. I had to pull the base of 10 out mechanically and actually put it into the equation. – user57055 Jan 12 '13 at 19:55
@user57055 : Mathematicians usually take $\log x$ to mean $\log_e x$. Some others often take it to mean $\log_{10} x$. – Michael Hardy Jan 12 '13 at 20:32

what are you guys talking about? $\log_{e}$ is written as $\ln$. The easiest way to solve this problem is simply writing logarithm in exponent form. First, move the $1/2$ up to an exponent (law of logs). Now we have $\log(x+2)^{1/2} = 2$. From there, (and it is base ten), rewrite in exponent form --> the base, raised to the answer, equals whats in from of the log. So $$10^2 = (x+2)^{1/2}$$ $$100 = \sqrt{x+2}$$ $$10000 = x+2$$ $$99998 = x$$

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See Michael Hardy's comment to the other answer, and note that the other answer covers both bases. Also, the last line has one too many 9. – epimorphic Dec 8 '14 at 5:02